# Distance spring decompresses when friction is involved

by becky_marie11
Tags: force, friction, kinetic energy, spring, work
 P: 8 1. The problem statement, all variables and given/known data A block of mass m = 4.30 kg slides along a horizontal table with speed v0 = 8.50 m/s. At x = 0 it hits a spring with spring constant k = 68.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by μ = 0.200. How far did the spring compress when the block first momentarily comes to rest? m=4.30 kg vo=8.50m/s k=68.00N/m μ=.200 x=? 2. Relevant equations Kf-Ki=W K=1/2mv2 Fs=-kx Ffric=μNf W=Fd 3. The attempt at a solution First I found the work of the whole system (friction and spring forces) by using Kf-Ki=W. For the initial kinetic energy, I used K=1/2mv2=155J. For the final kinetic energy, I used 0, because I'm measuring the distance after the mass came to a stop. So, plugging the numbers into the Work equation, I simply get W=-155J I'm pretty sure I'm on the right track so far? Then, using W=Fd (just multiplying the forcexdistance because both vectors are in the same direction) I plugged in: W=(-kx+μNf)x I work it out to a quadratic by plugging the numbers in... -68x2+8.44366x+155=0 I solve then for x (using the quadratic root finder on my calculator, so I'm not screwing up the quadratic..), but both of the answers it gives me are incorrect.. What am I doing wrong? Help please!!
Mentor
P: 41,466
 Quote by becky_marie11 Then, using W=Fd (just multiplying the forcexdistance because both vectors are in the same direction)
Careful: That only works when the force is constant. The spring force is not constant.

Hint: What's the energy stored in a compressed spring?

Also: Careful with signs. The work done on the mass by both the spring and the friction force is negative.
 Sci Advisor HW Helper Thanks P: 26,148 hi becky! welcome to pf! isn't it 1/2 kx2 ?
P: 33
Distance spring decompresses when friction is involved

 Quote by tiny-tim hi becky! welcome to pf! isn't it 1/2 kx2 ?
W = ∫Fdx (F dot dx, technically)
Fspring = kx

W = ∫kxdx
Wspring = kx2/2

Just to show where it came from, since W does NOT equal Fx, that is not a definition, it is a derivation of work for a constant force.
 P: 8 Thank you guys! I finally figured out the right answer! But when I solve for the quadratic equation, why is the distance vector only distributed to the friction force? This is what I tried to do... W=(-1/2kx^2+friction force)*x and this is what is right... W=-1/2kx^2+friction force*x Is it because the work of the spring is already in terms of "work" and the friction force isn't?